An aqueous solution freezes at $-2.55 \ ^oC$. What is its boiling point in $^oC$?
[$K_b(H_2O) = 0.52 \ K \ kg \ mol^{-1}, K_f(H_2O) = 1.86 \ K \ kg \ mol^{-1}$]

Benzene and naphthalene form an ideal solution at room temperature. For this process,the true statement$(s)$ is (are)
$(A)$ $\Delta G$ is positive
$(B)$ $\Delta S_{\text{system}}$ is positive
$(C)$ $\Delta S_{\text{surroundings}} = 0$
$(D)$ $\Delta H = 0$

Identify the false statement:

$A$ solution has a $1:4$ mole ratio of pentane to hexane. The vapour pressure of the pure hydrocarbons at $20 \ ^oC$ are $440 \ mm \ Hg$ for pentane and $120 \ mm \ Hg$ for hexane. The mole fraction of pentane in the vapour phase would be

In which solution,solute is liquid and solvent is gas?

For a solution formed by mixing liquids $L$ and $M$,the vapour pressure of $L$ plotted against the mole fraction of $M$ in solution is shown in the following figure. Here $x_L$ and $x_M$ represent mole fractions of $L$ and $M$,respectively,in the solution. The correct statement$(s)$ applicable to this system is(are)
$A$. Attractive intermolecular interactions between $L-L$ in pure liquid $L$ and $M-M$ in pure liquid $M$ are stronger than those between $L-M$ when mixed in solution
$B$. The point $Z$ represents vapour pressure of pure liquid $M$ and Raoult's law is obeyed when $x_L \rightarrow 0$
$C$. The point $Z$ represents vapour pressure of pure liquid $L$ and Raoult's law is obeyed when $x_L \rightarrow 1$
$D$. The point $Z$ represents vapour pressure of pure liquid $M$ and Raoult's law is obeyed from $x_L=0$ to $x_L=1$